# Easy Time and Work Solved QuestionAptitude Discussion

 Q. To complete a piece of work $A$ and $B$ take 8 days, $B$ and $C$ 12 days. $A$, $B$ and $C$ take 6 days. $A$ and $C$ will take :
 ✖ A. 7 Days ✖ B. 7.5 Days ✔ C. 8 Days ✖ D. 8.5 Days

Solution:
Option(C) is correct

Given $(A+B)$'s  one day’s work = $\dfrac{1}{8}$

$(B+C)$'s  one day’s work = $\dfrac{1}{12}$

$(A+B+C)$'s 1 day’s work = $\dfrac{1}{6}$

Work done by $A$, alone:

$=(A+B+C)$'s 1 day’s work - $(B+C)$'s  one day’s work

$= \dfrac{1}{6} - \dfrac{1}{12}$

$=\dfrac{1}{12}$

Work done by $C$, alone:

$=(A+B+C)$'s 1 day’s work - $(A+B)$'s  one day’s work

$= \dfrac{1}{6} - \dfrac{1}{8}$

$=\dfrac{1}{24}$

$(A+C)$'s one day’s work:

$=\dfrac{1}{12}+\dfrac{1}{24}$

$=\dfrac{1}{8}$

$(A+C)$ will take $\textbf{8 days}$ to complete the work together.

Edit:For an alternative solution, check comment by Manpreet Kunnath.

## (13) Comment(s)

Abhijeet Panchalwar
()

Take LCM then check the option which divisible by 24 LCM value I, E 8days

AAb
()

Please explain how to solve further

Crazy
()

By LCM method,

(A+B+C) (A+B) (B+C)

6 8 12 L.C.M=24

----------------------------

4 3 2

(A+B) +(B+C)+(C+A)= 2(A+B+C)= 2*4 units = 8 units

(A+B) +(B+C)+(C+A)= 8 units

(3 units)+(2 units)+ ((A+C) units)= 8 units

Therefore, A+C= 3 units

(A+B) (B+C) (A+C)

8 12 x L.C.M=24

----------------------------

3 2 3(i.e. units of A+C)

now, x=24/3= 8 days

ABHI
()

A+B = 1/8th of the work

B+C = 1/12th of the work

A+B+C = 1/6th of the work

C= 1/6-1/8 =1/24

A= 1/6- 1/12 = 1/12

C+A = 1/24+ 1/12 = 3/24 i.e., 1/8th of the work

So togather they will take 8days

Hope you guys know the basics

Pavan Kumar
()

Given (1/A+1/B)=1/8

(1/B+1/C)=1/12

(1/A+1/B+1/C)=1/6 Then

2* (1/A+1/B+1/C)=2/6 => (1/A+1/C)=2/6-1/8-1/12 =>

After solving above one we will get

(1/A+1/C)=1/8

so 8 days required to complete the work if A and C are worked to gether

Anvesh
()

How 1/A can be work done in one day ? Pls do reply

Manpreet Kunnath
()

$A+B+C=6$ ..... (1)

$A+B=8$ ..... (2)

Replacing (2) in (1)

$8+C=6$

Hence, $C=2$ (Ignoring the negative symbol)

$B+C=12$ ..... (3)

replacing (3) in (1)

$A+12=6$

Hence, $A=6$ (Ignoring the negative symbol)

$A+C = 6+2$

$= \textbf{8 days}$

Priyanshu Kumar
()

How can you ignore negation?

Its the luck which helped u here.

Suhail Ahmed
()

We have formula for this:

$=\dfrac{2XYZ}{XY+YZ+ZX}$

Simply solve for $=ZX$

S Malik
()

@Shireen like $A+B=8$ SO $A=8-B$ and so on

Shireen
()

Tried but could not work out the calculation, any help please.

Keep it simple. Put values of $A+B$ and $B+C$ in $A+B+C=6$ equation to get values of $C$ and $A$ simultaneously.